• # question_answer If$\cos \,(\theta -\alpha )=a,\,\,\sin \,(\theta -\beta )=b,\,\,$then ${{\cos }^{2}}(\alpha -\beta )+2ab\,\sin \,(\alpha -\beta )$ is equal to A) $4{{a}^{2}}{{b}^{2}}$ B) ${{a}^{2}}-{{b}^{2}}$ C) ${{a}^{2}}+{{b}^{2}}$ D) $-{{a}^{2}}{{b}^{2}}$

We have $\sin (\alpha -\beta )=\sin (\theta -\beta -\overline{\theta -\alpha })$ $=\sin (\theta -\beta )\cos (\theta -\alpha )-\cos (\theta -\beta )\sin (\theta -\alpha )$ $=ba-\sqrt{1-{{b}^{2}}}\sqrt{1-{{a}^{2}}}$ and $\cos (\alpha -\beta )=\cos (\theta -\beta -\overline{\theta -\alpha })$ $=\cos (\theta -\beta )\cos (\theta -\alpha )+\sin (\theta -\beta )\sin (\theta -\alpha )$ $=a\sqrt{1-{{b}^{2}}}+b\sqrt{1-{{a}^{2}}}$ $\therefore$ Given expression is ${{\cos }^{2}}(\alpha -\beta )+2ab\sin (\alpha -\beta )$ $=(a\sqrt{1-{{b}^{2}}}+b\sqrt{1-{{a}^{2}}{{)}^{2}}}+2ab\{ab-\sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}\}$ $={{a}^{2}}+{{b}^{2}}$. Trick : Put $\alpha =30{}^\circ ,\beta =60{}^\circ$ and $\theta =90{}^\circ ,$ then $a=\frac{1}{2},b=\frac{1}{2}$ $\therefore {{\cos }^{2}}(\alpha -\beta )+2ab\sin (\alpha -\beta )=\frac{3}{4}+\frac{1}{2}\times \left( -\frac{1}{2} \right)=\frac{1}{2}$ which is given by option (c).